Php 论形而上学与符号学中的一元关系
我有两个实体,叫做PictureTag和Tag,关系如下:Php 论形而上学与符号学中的一元关系,php,symfony,orm,doctrine-orm,Php,Symfony,Orm,Doctrine Orm,我有两个实体,叫做PictureTag和Tag,关系如下: /** * @ORM\Entity * @ORM\Table(name="picture_tag") * @ORM\HasLifecycleCallbacks() */ class PictureTag { /** * @var integer $id * * @ORM\Column(name="id", type="integer") * @ORM\Id * @ORM
/**
* @ORM\Entity
* @ORM\Table(name="picture_tag")
* @ORM\HasLifecycleCallbacks()
*/
class PictureTag
{
/**
* @var integer $id
*
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
*
* @ORM\OneToOne(targetEntity="App\MainBundle\Entity\Tag", inversedBy="id")
* @ORM\JoinColumn(name="tag_id", referencedColumnName="id", nullable=false)
*/
private $tag;
}
/**
* @ORM\Entity
* @ORM\Table(name="tag")
* @ORM\HasLifecycleCallbacks()
*/
class Tag
{
/**
* @var integer $id
*
* @ORM\Column(name="id", type="integer")
* @ORM\ManyToOne(targetEntity="App\MainBundle\Entity\PictureTag", inversedBy="tag")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @var string
* @ORM\Column(name="tag", type="string", nullable=true)
*/
private $tag;
}
基本上,我希望标记表包含所有唯一的标记,因此标记中没有重复项。我希望picturetag有一个joincolumn,指向标记的id。下面是我在控制器中的代码:
foreach ($image->tags as $tag) {
$existingTag = $em->getRepository('AppMainBundle:InstagramTag')->findOneByTag($tag);
$instaPictureTag = new PictureTag();
if ($existingTag) {
$instaPictureTag->setTag($existingTag);
} else {
$instagramTag = new Tag();
$instagramTag->setTag($tag);
$em->persist($instagramTag);
$instaPictureTag->setTag($instagramTag);
}
$instaPictureTag->setPicture($instaShopPicture);
$em->persist($instaPictureTag);
}
因此,这种关系基本上是:
One picture is going to have many tags. And one tag will belong to many pictures.
这里的图片标签是中间表
基本上,我正在检查标记是否已经存在于标记表中,如果已经存在,那么我将picturetag设置为与该标记关联,如果不存在,那么创建一个。但是,这样做会产生以下错误:
Fatal error: Call to a member function setValue() on a non-object in /Users/MyName/Sites/App/vendor/doctrine/orm/lib/Doctrine/ORM/UnitOfWork.php on line 2625
你知道为什么吗?据我所知,你想在PictureTag和Tag之间实现一个任意的双向关系 请尝试:
/**
* @ORM\Entity
* @ORM\Table(name="picture_tag")
* @ORM\HasLifecycleCallbacks()
*/
class PictureTag
{
/**
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\ManyToOne(targetEntity="DynamicSolutions\Bundle\ProsAndConsBundle\Entity\Tag", inversedBy="pictureTags")
* @ORM\JoinColumn(name="tag_id", referencedColumnName="id", nullable=false)
*/
private $tag;
}
/**
* @ORM\Entity
* @ORM\Table(name="tag")
* @ORM\HasLifecycleCallbacks()
*/
class Tag
{
/**
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\OneToMany(targetEntity="PictureTag", mappedBy="tag")
*/
private $pictureTags;
/**
* @var string
* @ORM\Column(name="tag", type="string", nullable=true)
*/
private $tag;
}
据我所知,您希望在PictureTag和Tag之间实现一对一的双向关系 请尝试:
/**
* @ORM\Entity
* @ORM\Table(name="picture_tag")
* @ORM\HasLifecycleCallbacks()
*/
class PictureTag
{
/**
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\ManyToOne(targetEntity="DynamicSolutions\Bundle\ProsAndConsBundle\Entity\Tag", inversedBy="pictureTags")
* @ORM\JoinColumn(name="tag_id", referencedColumnName="id", nullable=false)
*/
private $tag;
}
/**
* @ORM\Entity
* @ORM\Table(name="tag")
* @ORM\HasLifecycleCallbacks()
*/
class Tag
{
/**
* @ORM\Column(name="id", type="integer")
* @ORM\Id
* @ORM\GeneratedValue(strategy="AUTO")
*/
private $id;
/**
* @ORM\OneToMany(targetEntity="PictureTag", mappedBy="tag")
*/
private $pictureTags;
/**
* @var string
* @ORM\Column(name="tag", type="string", nullable=true)
*/
private $tag;
}